talaangoshtari wrote:In how many ways we can distribute 4 distinguishable books between 3 students in a way that each student receives at least one book?

A. 21
B. 25
C. 33
D. 36
E. 40

Since each student must receive at least 1 book, one of the 3 students must receive EXACLY 2 BOOKS.
Number of ways to choose a student to receive exactly 2 books = 3. (Any of the 3 students.)
From the 4 books, the number of ways to choose 2 to give to this student = 4C2 = (4*3)/(2*1) = 6.
Number of book options for the next student = 2. (Either of the 2 remaining books.)
Number of book options for the last student = 1. (Only 1 book left.)
To combine these options, we multiply:
3*6*2*1 = 36.

The correct answer is D.

talaangoshtari wrote:In how many ways we can distribute 4 distinguishable books between 3 students in a way that each student receives at least one book?

A. 21
B. 25
C. 33
D. 36
E. 40

Alternate approach:
WRITE OUT one case and use this result to calculate the remaining cases.

Let the 4 books be A, B, C and D.
Let student 1 receive exactly 2 books.
Options for distributing the books among the 3 students:
AB-C-D, AB-D-C --> 2 ways
AC-B-D, AC-D-B --> 2 ways
AD-B-C, AD-C-B --> 2 ways
BC-A-D, BC-D-A --> 2 ways
BD-A-C, BD-C-A --> 2 ways
CD-A-B, CD-B-A --> 2 ways.
Total ways if student 1 receives exactly 2 books = 2+2+2+2+2+2 = 12.

Since there are 12 ways if student 1 receives exactly 2 books, there must be 12 ways if student 2 receives exactly 2 books, and 12 ways if student 3 receives exactly 2 books.
Total ways = 12+12+12 = 36.

The correct answer is D.